Answer:
a= 2.7 m/s^2
Explanation:
acceleration: a
speed: V
Vf = final speed
Vi= initial speed (initial = beginning)
100 km/hour --> m/s
divide the speed value by 3.6
100/3.6= 27.8 m/s
a= 2.7 m/s^2
Answer:
Maximum speed of the car is 26.56 m/s.
Explanation:
Given that,
Mass of the car, m = 1200 kg
Radius of the curve, r = 90 m
The coefficient of friction between tires and the road is 0.8.
We need to find the maximum speed of the car. On the circular curve, the centripetal force is balanced by the force of friction. So,
So, the maximum speed of the car is 26.56 m/s.
Answer:
1.5x10^-4C
C=Couloumb's
Explanation:
Expression for the electric force between the two charges is given by -
F = (k*q1*q2) / r^2
Here, k = constant = 9 x 10^9 N*m^2 / C^2
F=25N
q1 = 1.9x10^-6 x 10^-6 C
q2 = ?
r = 0.32m
Substitute the given values in the above expression -
25N= 9x10^9 *1.9x10^-6 *q2 / 0.32m^2
25N= 17,100*q2 / 0.1024m
Next part is algebra multiply both sides by 0.1024 to remove denominator
2.56=17,100*q2
Divide both sides to isolate q2
q2= 1.5x10^-4C
Answer:
d. may be either greater, smaller, or equal to that observed inside the bus.
Explanation:
- The bus is moving at a constant speed. The ball tossed and received by the ball is inside the bus at a speed equal to the speed of the ball. Therefore the speed of the bus becomes zero with respect to the observer inside the bus.
- Now the observer inside the bus noticed the ball from the inside of the bus, so he threw the ball back and forth from the ball with the speed v relative to the observer.
- Now the observer outside the bus could see the bus moving at speed relative to its reference point and also throwing the ball from front to back. The speed of the ball to the observer outside the bus The speed of the bus to the observer outside the bus is minus the speed of the ball to the observer inside the bus.
- Therefore, the ball speed = (u-v) relative to the observer outside the bus.
